题目内容

已知数列{an}的前n项和Sn满足
an-1
Sn
=
a-1
a
(a>0),数列{bn}满足bn=an•logaan
(1)求数列{an}的通项;
(2)求数列{bn}的前n项和Tn
(1)当n=1时,a1=a>0且a≠1,
当n≥2时,Sn=
a
a-1
(an-1)
Sn-1=
a
a-1
(an-1-1)

两式相减得an=
a
a-1
(an-an-1)
,化为
an
an-1
=a

∴数列{an}是等比数列,an=an
(2)bn=an•lo
gana
=nan
当a=1时,Tn=1+2+…+n=
n(1+n)
2

当a≠1时,Tn=a+2a2+3a3+…+nan
aTn=a2+2a3+…++(n-1)an+nan+1
∴(1-a)Tn=a+a2+…+an-nan+1=
a(an-1)
a-1
-nan+1

∴Tn=
a+(na-n-1)an+1
(a-1)2
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